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Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
a: Ta có: \(x\left(x-3\right)-x^2+5=0\)
\(\Leftrightarrow-3x+5=0\)
hay \(x=\dfrac{5}{3}\)
b: Ta có: \(x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
\(a,\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
a: Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b: Ta có: \(x^2-5x-6=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
Vậy S = { 0, 6}
a. x(x-2)+x-2=0
=> (x-2).(x+1)=0
=> x-2=0 hoặc x+1=0
=> x=2 hoặc x=-1
b. 5x(x-3)-x+3=0
=> 5x(x-3)-(x-3)=0
=> (x-3).(5x-1)=0
=> x-3=0 hoặc 5x-1=0
=> x=3 hoặc x=1/5
`@` `\text {Ans}`
`\downarrow`
`a,`
`(x - 2)(x - 3) =0`
`<=>`\(\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0+2\\x=0+3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy, `S = {2; 3}`
`b,`
`x^2 - 5x = 0`
`<=> x(x - 5) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=0+5\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy, `S = {0; 5}`
`c,`
`x^2 - 9 = 0`
`<=> x^2 = 0 + 9`
`<=> x^2 = 9`
`<=> x^2 = (+-3)^2`
`<=> x = +-3`
Vậy, `S = {3; -3}`
`d,`
`4x^2 - 25 = 0`
`<=> 4x^2 = 25`
`<=> x^2 = 25/4`
`<=> x^2 = (+-5/2)^2`
`<=> x = +-5/2`
Vậy,` S = {5/2; -5/2}.`
a: =>x-2=0 hoặc x-3=0
=>x=2 hoặc x=3
b: =>x(x-5)=0
=>x=0 hoặc x=5
c: =>(x-3)(x+3)=0
=>x=3 hoặc x=-3
d: =>(2x-5)(2x+5)=0
=>x=5/2 hoặc x=-5/2
a) \(5x\left(x-1\right)=x-1\)
\(\Rightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right).\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(x=1\) hoặc \(x=\frac{1}{5}\)
b) \(2\left(x+5\right)-x^2-5x\)
\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)\)
\(\Rightarrow\left(x+5\right).\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
Vậy \(x=-5\)hoặc \(x=2\)
5x(x - 2) - 2 + x = 0
5x(x - 2) + (x - 2) = 0
(x - 2)(5x + 1) = 0
x - 2 = 0 hoặc 5x + 1 = 0
x = 2 hoặc x = \(\frac{-1}{5}\)